A company has five employees on its health insurance plan. Each year, each employee independently has an 80% probability of no hospital admissions. If an employee requires one or more hospital admissions, the number of admissions is modeled by a geometric distribution with a mean of 1.50. The numbers of hospital admissions of different employees are mutually independent. Each hospital admission costs 20,000.

Calculate the probability that the company’s total hospital costs in a year are less than 50,000.

• 0.41
• 0.46
• 0.58
• 0.69
• 0.78

Copyright 2023. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.

Solution: E

A geometric probability distribution with mean 1.5 will have p = 2/3. So Pr(1 visit) = 2/3, P(two visits) = 2/9, etc. There are four disjoint scenarios in which total admissions will be two or less.

Scenario 1: No employees have hospital admissions. Probability = 0.85 = 0.32768 .

Scenario 2: One employee has one admission and the other employees have none. Probability =

[[math]] \binom{5}{1}(0.2)(0.8)^4(2/3) = 0.27307. [[/math]]

Scenario 3: One employee has two admissions and the other employees have none. Probability =

[[math]] \binom{5}{1}(0.2)(0.8)^4(2/9) = 0.09102. [[/math]]

Scenario 4: Two employees each have one admission and the other three employees have none. Probability =

[[math]] \binom{5}{2} (0.2)^2(0.8)^3(2/3)(2/3) = 0.09102. [[/math]]

The total probability is 0.78279.

Copyright 2023 . The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.